Simplify. Multiply and remove all perfect squares from inside the square roots. Assume $y$ is positive. $\sqrt{12}\cdot\sqrt{y^3}\cdot\sqrt{6y}=$
Answer: Let's start by merging the square roots: $\begin{aligned} \sqrt{12}\cdot\sqrt{y^3}\cdot\sqrt{6y} &=\sqrt{12\cdot y^3\cdot 6y} \\\\ &=\sqrt{72y^4} \end{aligned}$ Now we remove all perfect squares from inside the square root: $\begin{aligned} \sqrt{72y^4} &=\sqrt{6^2\cdot 2\cdot \left(y^2\right)^2} \\\\ &=\sqrt{6^2}\cdot\sqrt{2}\cdot\sqrt{ \left(y^2\right)^2} \\\\ &=6\cdot \sqrt{2}\cdot y^2 \\\\ &=6y^2\sqrt{2} \end{aligned}$ In conclusion, $\sqrt{12}\cdot\sqrt{y^3}\cdot\sqrt{6y}=6y^2\sqrt{2}$